基于现有列自动生成新列答案

【问题标题】：Auto-generation of new columns based on existing columns基于现有列自动生成新列
【发布时间】：2020-06-30 13:07:54
【问题描述】：

我想根据 pandas 数据框中的其他列创建新列，并对其进行一些逻辑处理。对于名称为 [string] > Full Name 的每一列，我想生成两个名为 [string] > Asset Type 和 [string] > 的新列域。还有一些列没有这个[string]>Full Name结构，这些需要保持不变。

这是我所拥有的：

data = pd.DataFrame({'recipient > Full Name': {0: 'Norway', 1: 'Sweden'},
                    'transporter > Full Name': {0: "UPS", 1: "Sweden Mail Services"},
                    'Description': {0:'Priority mail', 1: 'Fragile object - be careful'}})

这就是我想要的：

wantedData = pd.DataFrame({'recipient > Full Name': {0: 'Norway', 1: 'Sweden'},
                    'transporter > Full Name': {0: "UPS", 1: "Sweden Mail Services"},
                    'Description': {0:'Priority mail', 1: 'Fragile object - be careful'},
                    'recipient > Asset Type': {0: "Country", 1: "Country"},
                    'recipient > Domain': {0: "Transport", 1: "Transport"},
                    'transporter > Asset Type': {0: "Legal Enitity", 1: "Legal Entity"},
                    'transporter > Domain': {0: "Transport", 1: "Transport"}})

此外，所有 Domain 列的所有行的值都相同，有没有办法用我在代码中使用的示例“Transport”自动填充它？

我尝试创建一些查看第 0 列并基于第 0 列创建第 1 列和第 2 列的代码 - 并迭代所有列，但这与我想要保持不变的列混淆。

【问题讨论】：

标签： python pandas automation transform

【解决方案1】：

import pandas as pd

data = pd.DataFrame({'recipient > Full Name': {0: 'Norway', 1: 'Sweden'},
                'transporter > Full Name': {0: "UPS", 1: "Sweden Mail Services"},
                'Description': {0:'Priority mail', 1: 'Fragile object - be careful'}})

# value_dict contains initial values for new columns (except those ending with Domain) 
value_dict = {
    'recipient > Asset Type' : 'Country',
    'transporter > Asset Type' : 'Legal Entity'
}

# key_list contains combination of new columns - in this case we want to create 
# 2 new columns (... Asset Type, ... Domain) for each column containing '>' 
key_list = ['Asset Type', 'Domain']

# iterate through list of column names containing '>' (other columns remain untouched)
for col in [col for col in data.columns if '>' in col]:
    # and for each such column create 2 new columns with new names (could be more or less...depends on key_list)
    for i in range(len(key_list)):
        new_colname = '{} > {}'.format(col.split(' >')[0], key_list[i%len(key_list)])
        # set Transport as value if column ends with '> Domain' or value from value_dict or None if not specified
        new_value = 'Transport' if new_colname.endswith('> Domain') else value_dict[new_colname] if new_colname in value_dict else None 
        data[new_colname] = new_value

输出：

【讨论】：